Question

If \[f(x) = xx - 1\], then\[\left( {fofo...of} \right)\left( x \right)19\] times is equal to:
A. \[xx - 1\]
B. \[\left( {xx - 1} \right)19\]
C. \[19xx - 1\]
D. \[x\]

Answer 1

Hint: If you can locate the inverse function of \[f\], you can solve for the missing value \[\left( x \right)\]. Take \[f\] 's derivative with respect to \[x\] and utilize substitution along with that knowledge to do this. You can accomplish this by determining the value of fo (or any other variable), adding it to the equation, and then changing back to x to obtain \[xofo...ofx = \dfrac{x}{{x - 1}}\].

Complete step by step solution: The given equation is \[f(x) = \dfrac{x}{{x - 1}}\]
In order to solve for \[x\] in this equation, we first need to simplify the negative divisor.
That is \[(fof)(x) = f\{ f(x)\} = f(\dfrac{x}{{x - 1}})\]
\[\begin{array}{l}\dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{x}{{x - 1}} - 1}} = \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{{x - x + 1}}{{x - 1}}}}\\ = > \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{1}{{x - 1}}}} = x\end{array}\]
So, \[(fofof)(x) = f(fof)(x) = f(x) = \dfrac{x}{{x - 1}}\]
\[(fofof...of)(x) = f(fof)(x) = f(x) = \dfrac{x}{{x - 1}}\]
If \[f\left( x \right)\]is a function that takes on one or more real numbers as input, and \[x\] occurs in the domain of \[f\left( x \right)\], then \[\left( {fofo...of} \right)\left( x \right)\]will always be equal to: \[\dfrac{x}{{x - 1}}\]. This property can be verified using integration by parts.
So, A is the answer.
That is , the \[(fofof...of)\]part is of \[19\] times which also gives the same value as it does not change much.
If \[f(x) = \dfrac{x}{{x - 1}}\], then \[(fofof...of)(x)\] \[19\]times is equal to: \[\dfrac{x}{{x - 1}}\]. That's because when we multiply two negative numbers together, the result will be a positive number. In this case, \[x\] gets multiplied by \[ - 1\]and that equals \[1\].

So, Option ‘A’ is correct.

Note: "If \[f\left( x \right) = \dfrac{x}{{x - 1}}\], then \[\left( {fofo...of} \right)\left( x \right)19\]times is equal to \[\dfrac{x}{{x - 1}}\]"
The equation's parentheses inform us that we are multiplying \[19\] terms together. Therefore, all we have to do to put the result of each phrase in parenthesis is place parentheses beneath the multiplication symbol and separate them.
The outcome of \[19\] times adding \[f\left( x \right)\]to itself is a real number in the range of \[ - 1\] and \[1\]. The distributive property can be used to determine this number.